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MATHEMATICS 339

Notes

MODULE - IICoordinateGeometry

Straight lines

10

STRAIGHT LINES

We have read about lines, angles and rectilinear figures in geometry. Recall that a line isthe join of two points in a plane continuing endlessly in both directions. We have alsoseen that graphs of linear equations, which came out to be straight lines

Interestingly, the reverse problem of the above is finding the equations of straight lines,under different conditions, in a plane. The analytical geometry, more commonly calledcoordinate geomatry, comes to our help in this regard. In this lesson. We shall find equationsof a straight line in different forms and try to solve problems based on those.

OBJECTIVES

After studying this lesson, you will be able to :

• derive equations of a line parallel to either of the coordinate axes;

• derive equations in different forms (slope-intercept, point-slope, two point, intercept,parametric and perpendicular) of a line;

• find the equation of a line in the above forms under given conditions;

• state that the general equation of first degree represents a line;

• express the general equation of a line into

(i) slope-intercept form (ii) intercept form and (iii) perpendicular form;

• derive the formula for the angle between two lines with given slopes;

• find the angle between two lines with given slopes;

• derive the conditions for parallelism and perpendicularity of two lines;

• determine whether two given lines are parallel or perpendicular;• derive an expression for finding the distance of a given point from a given line;• calculate the distance of a given point from a given line;

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Straight lines

• derive the equation of a line passing through a given point and parallel/perpendicular to agiven line;

• write the equation of a line passing through a given point and:

(i) parallel or perpendicular to a given line (ii) with given x-intercept or y-intercept(iii) passing through the point of intersection of two lines; and

• prove various geometrical results using coordinate geometry.

EXPECTED BACKGROUND KNOWLEDGE

• Congruence and similarity of traingles

10.1 STRAINGHT LINE PARALLEL TO AN AXIS

If you stand in a room with your arms stretched, we can have a line drawn on the floor parallelto one side. Another line perpendicular to this line can be drawn intersecting the first line betweenyour legs.

In this situation the part of the line in front of you and going behind you is they-axis and the onebeing parallel to your arms is the x-axis.

The direction part of the y-axis in front of you is positive and behind you is negative.

The direction of the part x-axis to your right is positive and to that to your left is negative.

Now, let the side facing you be at b units away from you, then the equation of this edge will bey = b (parallel to x-axis)

where b is equal in absolute value to the distance from thex-axis to the opposite side.

If b > 0, then the line lies in front of you, i.e., above thex-axis.

If b < 0, then the line lies behind you, i.e., below thex-axis.

If b = 0, then the line passes through you and is the x-axis itself.

Again, let the side of the right of you is atc units apart from you, then the equation of this line willbe x = c (parallel to y - axis)

where c is equal in absolute value, to the distance from they-axis on your right.

If c > 0, then the line lies on the right of you, i.e., to the right ofy-axis.

If c < 0, then the line lies on the left of you, i.e., to the left ofy-axis

If c = 0, then the line passes through you and is the y-axis.

Example 10.1 Find the equation of the lines passing through (2, 3) and is

(i) parallel to x-axis. (ii) parallel to y-axis.

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Solution :

(i) The equation of any line parallel to x-axis is y = b

Since it passes through (2, 3), hence b = 3

∴ The required equation of the line is y = 3

(ii) The equation of any line parallel to y-axis is x = c

Since it passes through (2, 3), hence c = 2

∴ The required equation of the line is x = 2.

Example 10.2 Find the equation of the line passing through (–2, –3) and

(i) parallel to x-axis (ii) parallel to y-axis

Solution :

(i) The equation of any line parallel to x-axis is y = b

Since it passes through (–2, –3), hence –3 = b

∴ The required equation of the line is y = –3

(i) The equation of any line parallel to y-axis is x = c

Since it passes through (–2, –3), hence –2 = c

∴ The required cquation of the line is x = –2

CHECK YOUR PROGRESS 10.1

1. If we fold and press the paper then what will the crease look like

2. Find the slope of a line which makes an angle of

(a) 45° with the positive direction of x-axis.

(b) 45° with the positive direction ofy-axis.

(c) 45° with the negative direction ofx-axis.

3. Find the slope of a line joining the points (2, –3) and (3, 4).

4. Determine x so that the slope of the line through the points (2,5) and (7,x) is 3.

5. Find the equation of the line passing through (–3, –4) and

(a) parallel to x-axis. (b) parallel to y-axis.

6. Find the equation of a line passing through (5, –3) and perpendicular to x-axis.

7. Find the equation of the line passing through (–3, –7) and perpendicular toy-axis.

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10.2 DERIVATION OF THE EQUATION OF STRAIGHT LINE IN VARIOUSSTANDARD FORMS

So far we have studied about the inclination, slope of a line and the lines parallel to the axes.Now the questions is, can we find a relationship betweenx and y, where (x, y) is any arbitrarypoint on the line?

The relationship betweenx and y which is satisfied by the co-ordinates of arbitrary point on theline is called the equation of a straight line. The equation of the line can be found in variousforms under the given conditions, such as

(a) When we are given the slope of the line and its intercept ony-axis.

(b) When we are given the slope of the line and it passes through a given point.

(c) When the line passes through two given points.

(d) When we are given the intercepts on the axes by the line.

(e) When we are given the length of perpendicular from origin on the line and the anglewhich the perpendicualr makes with the positive direction ofx-axis.

(f) When the line passes through a given point making an angleα with the positive directionof x-axis. (Parametric form).

We will discuss all the above cases one by one and try to find the equation of line in itsstandard forms.

(A) SLOPE-INTECEPT FORM

Let AB be a straight line making an angleθ with x-axis and cutting off an interceptOD =c from OY.

As the line makes intercept OD = c on y-axis, it is called y-intercept.

Let AB intersect OX' at T.

Take any point P(x, y) on AB. Draw PM ⊥ OX.

The OM = x, MP = y.

Draw DN ⊥ MP.

From the right-angled triangleDNP, we have

tan θ =NP

DN

MP MN

OM= –

=y OD

OM

–

=y c

x

– XX'

Y

Y'Fig. 10.1

c

O

ND

A

B

TM

x

y

P

θ

θ

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∴ y = x tan θ + c

tan θ = m (slope)

∴ y = mx + c

Since, this equation is true for every point on AB, and clearly for no other point in the plane,hence it represents the equation of the lineAB.

Note : (1) When c = 0 and m ≠ 0the line passes through the origin and its equation is y = mx

(2) When c = 0 and m = 0 ⇒ the line coincides with x – axis and its equation is of the form y = o

(3) When c ≠ 0 and m = 0 ⇒ the line is parallel to x-axis and its equation is of the form y = c

Example 10.3 Find the equation of a line with slope 4 and y-intercept 0.

Solution : Putting m = 4 and c = 0 in the slope intercept form of the equation, we get y = 4 x

This is the desired equation of the line.

Example 10.4 Determine the slope and the y-intercept of the line whose equation is

8x + 3y = 5.

Solution : The given equation of the line is 8x + 3y = 5

or, y = –8

3x + 5

3

Comparing this equation with the equationy = mx + c (Slope intercept form) we get

m = −8

3and c

5

3=

Therefore, slope of the line is –8

3 and its

y-intercept is5

3.

Example 10.5 Find the equation of the line cutting off an intercept of length 2 from the

negative direction of the axis ofy and making an angle of 1200 with the positive directionx-axis

Solution : From the slope intercept form of the line

∴ y = x tan 120° + (–2)

XX'

Y

Y'Fig. 10.2

2

O

A

B

0120

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Straight lines

= – 3 2x –

or, y + 3 2 0x + =

Here m = tan 120°, and c = –2, because the intercept is cut on the negative side of y -axis.

(b) POINT-SLOPE FORM

Here we will find the equation of a line passingthrough a given point A(x

1, y

1) and having the

slope m.

Let P(x, y) be any point other than A on given the

line. Slope (tan ) of the line joiningA(x

1, y

1) and P (x, y) is given by

m = tan =y y

x x

––

1

1

The slope of the line AP is given to be m.

∴ m =y y

x x

––

1

1

∴ The equation of the required line is

y – y1 = m (x – x

1)

Note : Since, the slope m is undefined for lines parallel to y-axis, the point-slopeform of the equation will not give the equation of a line though A (x

1, y

1) parallel to

y-axis. However, this presents no difficulty, since for any such line the abscissa ofany point on the line is x

1. Therefore, the equation of such a line is x = x

1.

Example 10.6 Determine the equation of the line passing through the point (2,– 1) and having

slope2

3.

Solution : Putting x1 = 2, y

1 = – 1 and m =

2

3 in the equation of the point-slope form of the line

we get

y – (– 1) =2

3 ( x – 2)

⇒ y + 1 =2

3 ( x – 2)

M

Fig. 10.3

θθ

1x x−

1y y−

1( )P x y

1 1( , )x y

O

A

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⇒ y =2

3 x –

7

3

which is the required equation of the line.

(c) TWO POINT FORM

Let A(x1, y

1) and B(x

2, y

2) be two given distinct points.

Slope of the line passing through these points is given by

m =y y

x x2 1

2 1

–– (x

2 ≠ x1)

From the equation of line in point slope form, we get

2 11 1

2 1

( )y y

y y x xx x

−− = −−

which is the required equation of the line in two-point form.

Example 10.7 Find the equation of the line passing through (3, – 7) and (– 2,– 5).

Solution : The equation of a line passing through two points (x1, y

1) and (x

2, y

2) is given by

y –y1 =

y y

x x2 1

2 1

−− (x – x

1) (i)

Since x1 = 3, y

1 = – 7 and x

2 = – 2, and y

2 = – 5, equation (i) becomes,

y + 7 =–– –

5 7

2 3

+ (x –3)

or, y + 7 =2

5– (x –3)

or, 2 x + 5 y + 29 = 0

(d) INTERCEPT FORM

We want to find the equation of a line whichcuts off given intercepts on both theco-ordinate axes.

Let PQ be a line meeting x-axis in A andy-axis in B. Let OA = a, OB = b.

Then the co-ordinates of A and B are(a,0) and (0, b,) respectively.

Fig. 10.4

XX'

Y

Y'

O

b

aA (a, )0

B ( ,b)0

P

Q

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The equation of the line joiningA and B is

y – 0 =b

a

––

0

0 (x – a)

or, y = –b

a (x – a)

or,y

b= –

x

a + 1

or,x

a +

y

b = 1

This is the required equation of the line having interceptsa and b on the axes.

Example 10.8 Find the equation of a line which cuts off intercepts 5 and –3 on x and y axes

respectively.

Solution : The intercepts are 5 and –3 on x and y axes respectively. i.e., a = 5, b = – 3

The required equation of the line is

15 3

x y+ =−

3 x – 5y –15 = 0

Example 10.9 Find the equation of a line which passes through the point (3, 4) and makes

intercepts on the axes equl in magnitude but opposite in sign.

Solution : Let the x-intercept and y-intercept be a and –a respectively

∴ The equation of the line is

1x y

a a+ =

−

x –y = a (i)

Since (i) passes through (3, 4)

∴ 3 – 4 = a or

or, a = –1

Thus, the required equation of the line is

x – y = – 1

or x – y + 1 = 0

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Example 10.10 Determine the equation of the line through the point (– 1,1) and parallel to x - axis.

Solution : Since the line is parallel to x-axis its slope ia zero. Therefore from the point slopeform of the equation, we get

y – 1 = 0 [ x – (– 1)]

y – 1 = 0

which is the required equation of the given line

Example 10.11 Find the intercepts made by the line

3x –2y + 12 = 0 on the coordinate axes

Solution : Equation of the given line is

3x – 2y = – 12.

Dividing by – 12, we get

x y

–4 6+ = 1

Comparing it with the standard equation of the line in intercept form, we finda = –4 and b =6. Hence the intercepts on the x-axis and y-axis repectively are –4. and 6.

Example 10.12 The segment of a line, intercepted between the coordinate axes is

bisected at the point (x1, y

1). Find the equation of the line

Solution : Let P(x1, y

1) be the middle point or the

segment CD of the line AB intercepted between theaxes. Draw PM OX⊥

∴ OM = x1 and MP = y

1

∴ OC = 2x1 and OD = 2y

1

Now, from the intercept form of the line

x

x2 1 +

y

y2 1 = 1

or,x

x1 +

y

y1 = 2

which is the required equation of the line.

Fig. 10.5

XX'

Y

Y'

O

D

1 1( , )P x y

B

M1x1y

C

A

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Straight lines

(e) PERPENDICULAR FORM (NORMAL FORM)

We now derive the equation of a line when p be the length of perpendicular from the origin onthe line and , the angle which this perpendicular makes with the positive direction ofx-a x is are given.

XX'

Y

Y'

O

1 1( , )P x y

b

a

(i)

B

A

P

pXX'

Y

Y'

Oa

(ii)

bp

180-

B

A

Fig. 10.6

Fig. 10.6

(i) Let A B be the given line cutting off interceptsa and b on x–axis and y –axis respectively.Let OP be perpendicular from origin O on A B and ∠ POB = (See Fig. 10.6 (i))

∴p

a = cos ⇒ a = p sec

and,p

b = sin⇒ b = p cosec

∴ The equation of line AB is

1sec cos

x y

p p ec + =

or, x cos + y cosec = p

(ii)p

a = cos (180° – ) = – cos [From Fig. 10.6 (ii)]

⇒ a = –p sec

similary, b = p cosec

∴ The equation of the line AB isx

a

y

b–+ = 1

x cos + y sin = p

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Note : 1. p is the length of perpendicular from the origin on the line and in alwaystaken to be positive.

2. is the angle between positive direction of x-axis and the line perpendicularfrom the origin to the given line.

Example 10.13 Determine the equation of the line with = 135° and perpendicular distance

p = 2 from the origin.

Solution : From the standard equation of the line in normal form have

x cos 135° + y sin 135° = 2

or, – x y

2 22+ =

or, –x + y – 2 = 0

or, x – y + 2 = 0

which is the required equation of the straight line.

Example 10.14 Find the equation of the line whose perpendicular distance from the origin is

6 units and the perpendicular from the origin to line makes an angle of 30° with the positive

direction of x-axis.

Solution : Here = 30°, p = 6


x cos 30° + y sin 30° = 6

or, x y3

2

1

26

FHGIKJ +FHGIKJ =

or, 3 12x y+ =

(f) PARAMETRIC FORM

We now want to find the equation of a linethrough a given point Q (x

1, y

1) which makes

an angle with the positive direction ofx – axis in the form

1 1

cos sin

x x y yr

− −= = XX'

Fig. 10.7x

O

B

Ny

P(x,y)

ML

1 1( )Q x y

r

A

1y

1x

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Straight lines

where r is the distance of any point P (x, y) on the line from Q (x1, y

1).

Let AB be a line passing through a given point Q (x1, y

1) making an angle with the positive

direction of x-axis.

Let P (x, y) be any point on the line such that QP = r (say)

Draw QL, PM perpendiculars to x-axis and draw QN ⊥ PM.

From right angled ∆ PNQ

cos =QN

QP , sin =PN

QP

But QN = LM = OM – OL = x − x1

and PN = PM – MN = PM – QL = y – y1

∴ cos =x x

r

– 1 , sin =y y

r

– 1

or, 1 1 ,cos sin

x x y yr

− −= = which is the required equation of the line

Note : 1. The co-ordinates of any point on this line can be written as(x

1 + r cos , y

1 + r sin ). Clearly coordinates of the point depend on the value of

r. This variable r is called parameter.

2. The equation x = x1 + r cos , y = y

1 + r sin are called the parametric equations

of the line.

3. The value of 'r' is positive for all points lying on one side of the given point andnegative for all points lying on the other side of the given point.

Example 10.15 Find the equation of a line passing through A (–1, –2) and making an angle

of 30° with the positive direction ofx-axis, in the parametric form. Also find the coordinates of

a point P on it at a distance of 2 units from the point A.

Solution : Here x1 = –1, y

1 = – 2 and = 30°


x y+°

= +°

1

30

2

30cos sin

or,

x y+ = +1

32

212

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Since AP = 2, r = 2

∴

x y+ = + =1

32

212

2

⇒ x = 3 – 1, y = – 1

Thus the coordinates of the point P are 3 1 1– , –d i Example 10.16 Find the distance of the point (1, 2) from the line 2x – 3y + 9 = 0

measured along a line making an angle of 45° withx – axis.

Solution : The equation of any line through A (1, 2) making an angle of 45° withx-axis is

x yr

–cos

–sin

1

45

2

45°=

°=

x yr

– –11

2

21

2

= =

Any point on it is 12

22

+ +FHG

IKJ

r r,

It lies on the given line 2x – 3y + 9 = 0

if 2 12

3 22

9 0+FHGIKJ +FHG

IKJ + =r r–

or, 2 6 92

0– –+ =r

or, r = 5 2

Hence the required distance is 5 2 .

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Straight lines


1. (a) Find the equation of a line with slope 2 and y – intercept is equal to –2.

(b) Determine the slope and the intercepts made by the line on the axes whose equationis 4x + 3y = 6.

2. Find the equation of the line cutting off an interecept1

3 on negative direction of axis of

y and inclined at 120° to the positive direction ofx-axis.

3. Find the slope and y-intercept of the line whose equation is 3x – 6y = 12.

4. Determine the equation of the line passing through the point (–7, 4) and having the slope

– .3

7

5. Determine the equation of the line passing through the point (1, 2) which makes equalangles with the two axes.

6. Find the equation of the line passing through (2, 3) and parallel to the line joining thepoints (2, –2) and (6, 4).

7. (a) Determine the equation of the line through (3, –4) and (–4, 3).

(b) Find the equation of the diagonals of the rectangle ABCD whose vertices areA (3, 2), B (11, 8), C (8, 12) and D (0, 6).

8. Find the equation of the medians of a triangle whose vertices are (2, 0), (0, 2) and (4, 6).

9. Find the equation of the line which cuts off intercepts of length 3 units and 2 units onx-axis and y-axis respectively.

10. Find the equation of a line such that the segment between the coordinate axes has its midpoint at the point (1, 3)

11. Find the equation of a line which passes through the point (3, –2) and cuts off positiveintercepts on x and y axes in the ratio of 4 : 3.

12. Determine the equation of the line whose perpendicular from the origin is of length 2 unitsand makes an angle of 45° with the positwe direction ofx-axis.

13. If p is the length of the perpendicular segment from the origin, on the line whose intercepton the axes are a and b, then show that

1 1 12 2 2p a b

= + .

14. Find the equation of a line passing through A (2, 1) and making an angle 45° with thepositive direction of x-axis in parametric form. Also find the coordinates of a pointP onit at a distance of 1 unit from the point A.

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10. 3 GENERAL EQUATION OF FIRST DEGREE

You know that a linear equation in two variables x and y is given by Ax + B y + C = 0 ...(1)

In order to understand its graphical representation, we need to take the following thres cases.

Case-1: (When both A and B are equal to zero)

In this case C is automaticaly zero and the equation does not exist.

Case-2: (When A = 0 and B ≠ 0)

In this case the equation (1) becomes By + C = 0.

or y = – C

Band is satisfied by all points lying on a line which is parallel to x-axis and the

y-coordinate of every point on the line is – C

B. Hence this is the equation of a straight line. The

case where B = 0 and A ≠ 0 can be treated similarly..

Case-3: (When A ≠ 0 and B ≠ 0)

We can solve the equation (1) for y and obtain.

yA

Bx

C

B= – –

Clearly, this represents a straight line with slope – A

B and y −intercept equal to – C

B.

10.3.1 CONVERSION OF GENERAL EQUATION OF A LINE INTO VARIOUSFORMS

If we are given the general equation of a line, in the form Ax + By + C = 0, we will see how thiscan be converted into various forms studied before.

10.3.2 CONVERSION INTO SLOPE-INTERCEPT FORM

We are given a first degree equation in x and y as Ax + By + C = O

Are you able to find slope and y-intercept ?

Yes, indeed, if we are able to put the general equation in slope-intercept form. For this purpose,let us re-arrange the given equation as.

Ax + By + C = 0 as

By = – Ax – C

or y = – –A

Bx

C

B (Provided B ≠ 0)

which is the required form. Hence, the slope = – A

B, y – intercept = – C

B.

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Example 10.17 Reduce the equation x + 7y – 4 = 0 to the slope – intercept form.

Solution : The given equation is

x + 7y – 4 = 0

or 7y = – x + 4

or y = +– 1

7

4

7x

which is the converted form of the given equation in slope-intercept form

Example 10.18 Find the slope and y intercept of the line x + 4y – 3 = 0.


x + 4y – 3 = 0

or 4y = – x + 3

or y = – 1

4

3

4x +

Comparing it with slope-intercept form, we have

slope = – 1

4, y – intercept =

3

4.

10.3.3 CONVERSION INTO INTERCEPT FORM

Suppose the given first degree equation in x and y is

Ax + By + C = 0. ...(i)

In order to convert (i) in intercept form, we re arrange it as

Ax + By = –C orAx

C

By

C– –+ = 1

or

xCA

yCB

(– ) (– )+ = 1

(Provided A ≠ 0 and B ≠ 0)

which is the requied converted form. It may be noted that intercept on x – axis =–CA and

intercept on y – axis =–CB

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Example 10.19 Reduce 3x + 5 y = 7 into the intercept form and find its intercepts on the

axes.


3x + 5y = 7

or,3

7

5

71x y+ =

or,

x y73

75

1+ =

∴ The x– intercept =7

3and, y – intercept =

7

5

Example 10.20 Find x and y-intercepts for the line 3x –2y = 5.


3x – 2y = 5

or,3

5

2

51x y– =

or,1

5 53 2

x y+ =−

Thus, the required x – intercept =5

3

and y-intercept = –5

2

10.3.4 CONVERSION INTO PERPENDICULAR FORM

Let the general first degree equation in x and y be

Ax + By + C = 0 ... (i)

We will convert this general equation in perpendicular form. For this purpose let us re-write thegiven equation (i) as Ax + By = – C

Multiplying both sides of the above equation byλ, we have

λAx + λBy = – λC ... (ii)

Let us choose λ such that (λA)2 + (λB)2 = 1

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or λ =1

2 2( )A B+ (Taking positive sign)

Substituting this value ofλ in (ii), we have

AxA B

ByA B

CA B( ) ( )

–( )2 2 2 2 2 2+

++

=+ ... (iii)

This is required conversion of (i) in perpendicular form. Two cases arise according as C isnegative or positive.

(i) If C < 0, the equation (ii) is the required form.

(ii) If C > 0, the R. H. S. of the equation of (iii) is negative.

∴ We shall multiply both sides of the equation of (iii) by –1.

∴ The required form will be

–( )

–( ) ( )

Ax

A B

By

A B

C

A B2 2 2 2 2 2+ +=

+

Thus, length of perpendicular from the origin =| |

( )

C

A B2 2+

Inclination of the perpendicular with the positwe direction of x-ax is

is given by cos θ = A

A B2 2+

or sin θ = B

A B( )2 2+

FHG

IKJ

where the upper sign is taken for C > 0 and the lower sign for C < 0. If C = 0, the linepasses through the origin and there is no perpendicular from the origin on the line.

With the help of the above three cases, we are able to say that

"The general equation of first degree in x and y always represents a straightline provided A and B are not both zero simultaneously."

Is the converse of the above statement true? The converse of the above statement is thatevery straight line can be expressed as a general equation of first degree in x and y.

MATHEMATICS 357

Notes


Straight lines

In this lesson we have studied about the various forms of equation of straight line. For example,

let us take some of them as y = mx + c,x

a

y

b+ = 1 and x cos α + y sin α = p. Obviously, all

are linear equations in x and y. We can re-arrange them as y – mx – c = 0, bx + ay – ab = 0and x cos α + y sin α – p = 0 respectively. Clearly, these equations are nothing but a differentarrangement of general equation of first degree inx and y. Thus, we have established that

"Every straight line can be expressed as a general equation of first degree in xand y".

Example 10.21 Reduce the equation x + 3 y + 7 = 0 into perpendicular form.

Solution : The equation of given line is x + 3y + 7 = 0 ... (i)

Comparing (i) with general equation of straight line, we have

A = 1 and B = 3

∴ 2 2 2A B+ =

Dividing equation (i) by 2, we have

xy

2

3

2

7

20+ + =

or – – –1

2

3

2

7

20F

HGIKJ +FHGIKJ =x y

or x cos4

3

+ y sin

4

3

=

7

2

(cos θ and sinθ being both negative in the third quadrant, value ofθ will lie in the third quadrant).

This is the representation of the given line in perpendicular form.

Example 10.22 Find the perpendicular distance from the origin on the line

3 x – y + 2 = 0. Also, find the inclination of the perpendicular from the origin.

Solution : The given equation is 3 x – y + 2 = 0

Dividing both sides by ( ) (–1)3 2 2+ or 2, we have

MATHEMATICS

Notes


358

Straight lines

3

2

1

21 0x y– + =

or,32

12

1x y– –=

Multiplying both sides by –1, we have

– 3

2

1

21x y+ =

or, x cos5

6

+ y sin

5

6

= 1 (cos θ is –ve in second quadrant and sin θ is +ve in second

quadrant, so value of θ lies in the second quadrant).

Thus, inclination of the perpendicular from the origin is 150° and its length is equal to 1.

Example 10.23 Find the equation of a line which passess through the point (3,1) and bisects

the portion of the line 3x + 4y = 12 intercepted between coordinate axes.

Solution : First we find the intercepts on coordinate axes cut off by the line whose equation isfrom. 3x + 4y = 12

or3

12

4

121

x y+ =

orx y

4 31+ =

Hence, intercepts on x-axis and y-axis are 4 and 3 respectively.

Thus, the coordinates of the points where the line meets the coordinate axes areA (4, 0) andB (0, 3).

∴ Mid −point of AB is 23

2,FHGIKJ .

Hence the equation of the line through (3, 1)

and 23

2,FHGIKJ is

y – 1 =

32

1

2 33

–

–( – )x

x

Fig. 10.8

O

B

A(4,0)(3,1)

C

D(0,3)

y

32,

2

MATHEMATICS 359

Notes


Straight lines

or y x– ( – )11

23= −

or 2 (y –1) + (x – 3) = 0

or 2y – 2 + x – 3 = 0

or x + 2y – 5 = 0

Example 10.24 Prove that the line through (8, 7) and (6, 9) cuts off equal intercepts on

coordinate axes.

Solution : The equation of the line passing through (8, 7) and (6, 9) is

y x– ––

( – )79 7

6 88=

or y – 7 = – (x – 8)

or x + y = 15

orx y

15 151+ =

Hence, intercepts on both axes are 15 each.

Example 10.25 Find the ratio in which the line joining (–5, 1) and (1, –3) divides the join of(3, 4) and (7, 8).

Solution : The equation of the line joining C (–5, 1) and D (1, –3) is

y x– –3 –( )1

1

1 55=

++

or4

1 ( 5)6

y x− = − +

or 3y – 3 = –2x – 10

or 2x + 3y + 7 = 0 ... (i)

Let line (i) divide the join of A (3, 4) and B (7, 8) atthe point P.

If the required ratio is λ : 1 in which line (i) dividesthe join of A (3, 4) and B (7, 8), then the coordinatesof P are

7 3

1

8 4

1

++

++

FHG

IKJ,

X

Fig. 10.9

P(x,y)

B(7,8)

Y

A(3,4)

o

( )5,1C −

( )1, 3D −

MATHEMATICS

Notes


360

Straight lines

Since P lies on the line (i), we have

27 3

13

8 4

17 0

++FHGIKJ +

++FHGIKJ + =

⇒ 14λ + 6 + 24λ + 12 + 7λ + 7 = 0

⇒ 45λ + 25 = 0 ⇒ λ = – 5

9

Hence, the line joining (–5, 1) and (1, –3) divides the join of (3, 4) and (7, 8) externally in theratio 5 : 9.


1. Under what condition, the general equation Ax + By + C = 0 of first degree in x and yrepresents a line?

2. Reduce the equation 2x + 5y + 3 = 0 to the slope intercept form.

3. Find the x and y intercepts for the following lines :

(a) y = mx + c (b) 3y = 3x + 8 (c) 3x – 2y + 12 = 0

4. Find the length of the line segment AB intercepted by the straight line 3x – 2y + 12 = 0between the two axes.

5. Reduce the equation x cos + y sin = p to the intercept form of the equation andalso find the intercepts on the axes.

6. Reduce the following equations into normal form.

(a) 3x – 4y + 10 = 0 (b) 3x – 4y = 0

7. Which of the lines 2x – y + 3 = 0 and x – 4y – 7 = 0 is nearer from the origin?

We will here develop a formula for finding angle between two given lines

10.4 ANGLE BETWEEN TWO LINES

We will now try to find out the angle between the arms of the divider or clock when equationof the arms are known. Two methods are discussed below.

10.4.1 When two lines with slopes m1 and m

2are given

Let AB and CD be two straight lines whose equations are

y = m1

x + c1 and y = m

2x + c

2 respectively.

Let P be the point of intersection of AB and CD.

L e t ∠XEB = θ1 and ∠XFD = θ2. Then

MATHEMATICS 361

Notes


Straight lines

tan θ1 = m1 and tan θ2 = m

2

Let θ be the angle ∠EPF between AB and CD.

Now θ1 = θ2 + θ

or θ = θ1 – θ2

∴ tan θ = tan (θ1 – θ2)

or tan θ =tan – tan

tan tan

1 2

1 21 +

Hence, tan θ = m m

m m1 2

1 21

–+

which gives tangent of the angleθ between two givenlines in terms of their slopes.

From this result we can find the angle between any two given lines when their slopes are given.

Note : (i)21

1 2

0,1

m mIf

m m

−>

+ then angle between lines is acute.

(ii)21

1 2

0,1

m mIf

m m

−<

+ then angle between lines is obtuse.

(iii)21

1 2

0,1

m mIf

m m

−=

+

i.e., m1 = m

2, then lines are either coincident or parallel.

i.e, Two lines are parallel if they differ only by a constant term.

(iv) If the denominator ofm m

m m1 2

1 21

–+ is zero.

i.e, m1m

2 = –1, then lines are perpendicular.

Example 10.26 Find the angle between the lines 7x – y = 1 and 6x – y = 11.

Solution : Let θ be the angle betwen the lines 7x – y = 1 and 6x – y = 11. The equations ofthese lines can be written as y = 7 x – 1 and y = 6x – 11.

Here, m1= 7, m

2 = 6. where m

1and m

2 are respective slopes of the given lines.

∴ tan θ =7 6

1 42

1

43

–+

=

Fig. 10.10

D

xx'

y

y'

O P

B

C A

OEF

1θ2θ

θ

MATHEMATICS

Notes


362

Straight lines

The angle θ between the given lines is given by tan θ =1

43

Example 10.27 Find the angle between the lines x + y + 1 = 0 and 2x – y – 1 = 0.

Solution : Putting the equations of the given lines in slope intercept form, we get y = –x – 1and y = 2x – 1

∴ m1 = – 1 and m

2 = 2 where m

1, and m

2 are respectively slopes of the given lines.

∴ tan θ =–1 –(–1) ( )

2

1 23

+=

The angle θ between the given lines is given by tan θ = 3

Example 10.28 Are the straight lines y = 3x – 5 and 3x + 4 = y, parallel or perpendicular?

Solution : For the first line y = 3x – 5, the slope m1 = 3 and for the second line 3x + 4 = y,

the slope m2 = 3. Since the slopes of both the lines are same for both the lines, hence the

lines are parallel.

Example 10.29 Are the straight lines y = 3x and y = – 1

3x , parallel or perpendicular?

Solution : The slope m1 of the line y = 3x is 3 and slope m

2 of the second line y = – x

3 is – 1

3.

Since, m1m

2 = –1, the lines are perpendicular.

10.4.2 When the Equation of two lines are in general form

Let the general equations of two given straight lines be given by

A1

x + B1

y + C1 = 0 ... (i)

and A2

x + B2

y + C2 = 0 ...(ii)

Equations (i) and (ii) can be written into their slope intercept from as

yA

Bx

C

B= – –1

1

1

1 and y

A

Bx

C

B= – –2

2

2

2

Let m1 and m

2 be their respective slopes and using the formula, we get.

i.e., tan– =+

A B A B

A A B B2 1 1 2

1 2 1 2

Thus, by using this formula we can calculate the angle between two lines when they are ingeneral form.

MATHEMATICS 363

Notes


Straight lines

Note : (i) If A2

B1 – A

1B

2 = 0

or A1

B2 – A

2B

1 = 0

i.e.A B

A B1 1

2 2 = 0, then the given lines are parallel.

(ii) If A1A

2 + B

1B

2 = 0, then the given lines are perpendicular.

Example 10.30 Are the straight lines x – 7y + 12 = 0 and x – 7 y + 6 = 0, parallel?

Solution : Here A = 1, B1 = – 7; A

2 = 1 and B

2 = – 7

SinceA B

A B1 1

2 2 =1 71 7–– = 0

Example 10.31 Determine whether the straight lines 2x + 5y – 8 = 0 and 5x – 2y – 3 = 0 areparallel or perpendicular ?

Here A1 = 2, B

1 = 5

A2 = 5 and B

2 = – 2

Solution : Here A1

A2 + B

1B

2 = (2) × (5) + (5) × (–2) = 0.

Hence the given lines are perpendicular.


1. What is the condition for given lines y = m1 x + c

1 and Ax + By + C = 0

(a) to be parallel? (b) to be perpendicular?

2. Find the angle between the lines 4x + y = 3 andx

2 + y =

4

7.

3. (a) Write the condition that the two lines A1 x + B

1 y + C

1 = 0 and A

2 x + B

2y + C

2 = 0

are

(i) parallel.

(ii) perpendicular.

(b) Are the straight lines x – 3y = 7 and 2x – 6y –16 = 0, parallel?

(c) Are the straight lines x = y + 1 and x = –y + 1 perpendicular?

MATHEMATICS

Notes


364

Straight lines

10.5 DISTANCE OF A GIVEN POINT FROM A GIVEN LINE

In this section, we shall discuss the concept of finding the distance of a given point from a givenline or lines.

Let P(x1, y

1) be the given point and l be the line Ax + By + C = 0.

Let the line l intersect x axis and y axis R and Q respectively.

Draw PM ⊥ l and let PM = d.

Let the coordinates of M be (x2, y

2)

d = x x y y1 2

2

1 2

2– –b g b g }{ + ...(i)

M lies on l

∴ Ax2 + By

2 + C = 0

or C = – (Ax2 + By

2) ...(ii)

The coordinates of R and Q are – ,C

A0F

HGIKJ and 0, – C

BFHGIKJ respectively..

The slope of QR =

0

0

+=

C

BC

A

A

B– –– and,

the slope of PM =2 1

2 1

y y

x x

−−

As PM ⊥ QR ⇒ 2 1

2 1

1.y y A

x x B

− × − = − −

ory y

x x

B

A1 2

1 2

––

= ...(iii)

Fig. 10.11

xx'

y

y'

O

1 1( )P x y

M

d

R

Q

MATHEMATICS 365

Notes


Straight lines

From (iii)x x

A

y y

B

x x y y

A B

1 2 1 21 2

2

1 2

2

2 2

– – – –= =

+

+

b g b g }{c h ...(iv)

(Using properties of Ratio and Proportion)

Alsox x

A

y y

B

A x x B y y

A B1 2 2 1 2 1 2

2 2

– – – –= =

++

b g b g...(v)

From (iv) and (v), we get

x x y y

A B

A x x B y y

A B

1 2

2

1 2

2

2 2

1 2 1 22 2

– – – –b g b g }{c h

b g b g+

=++

ord

A B

Ax By Ax By

A B2 2

1 1 2 22 2+

+ + ++

– ([Using (i)]

orAx By C

A B

1 1

2 2

+ +

+c h [Using (ii)]

Since the distance is always positive, we can write

d =Ax By C

A B

1 1

2 2

+ +

+c hNote : The perpendicular distance of the origin (0, 0) from Ax + By + C = 0 is

A B C

A B

( ) ( )0 02 2

+ +

+c h ( )2 2

C

A B=

+

Example 10.32 Find the points on the x-axis whose perpendicular distance from the straight

linex

a

y

b+ = 1 is a.

Solution : Let (x1, 0) be any point on x-axis.

Equation of the given line is bx + ay – ab = 0. The perpendicular distance of the point (x1, 0)

from the given line is

MATHEMATICS

Notes


366

Straight lines

abx a ab

a b= ± +

+1

2 2

0. –

c h

∴ xa

bb a b1

2 2= ± +RST c h}

Thus, the point on x-axis isa

bb a b± +FHG

IKJ

2 2 0e j ,


1. Find the perpendicular distance of the point (2, 3) from 3x + 2y + 4 = 0.

2. Find the points on the axis of y whose perpendicular distance from the straight line

x

a

y

b+ = 1 is b.

3. Find the points on the axis of y whose perpendicular distance from the straight line 4x +3y = 12 is 4.

4. Find the perpendicular distance of the origin from 3x + 7y + 14 = 0

5. What are the points on the axis of x whose perpendicular distance from the straight line

4 4

x y+ = 1 is 3?

10.6 EQUATION OF PARALLEL (OR PERPENDICULAR) LINES

Till now, we have developed methods to find out whether the given lines are prallel orperpendicular. In this section, we shall try to find, the equation of a line which is parallel orperpendicular to a given line.

10.6.1 EQUATION OFA STRAIGHT LINE PARALLELTO THE GIVEN LINE

Ax + By + c = 0

Let A1x + B

1y + C

1 = 0 ...(i)

be any line parallel to the given line

Ax + By + C = 0

The condition for parallelism of (i) and (ii) is ...(ii)

A

A

B

B1 1= = K

1(say)

⇒ A1 = AK

1, B

1 = BK

1

MATHEMATICS 367

Notes


Straight lines

with these values of A1 and B

1, (i) gives

AK1x + BK

1 y + C

1 = 0

or Ax + By +C

K1

1 = 0

or Ax + By + K = 0, where K =C

K1

1... (iii)

This is a line parallel to the given line. From equations (ii) and (iii) we observe that

(i) coefficients of x and y are same

(ii) constants are different, and are to evaluated from given conditions.

Example 10.33 Find equation of the straight line, which passes through the point (1, 2) and

which is parallel to the straight line 2x + 3y + 6 = 0.

Solution : Equation of any straight line parallel to the given equation can be written if we put

(i) the coefficients of x and y as same as in the given equation.

(ii) constant to be different from the given equation, which is to be evaluated under givencondition.

Thus, the required equation of the line will be

2x + 3y + K = 0 for some constant K

Since it passes through the point (1, 2) hence

2 × 1 + 3 × 2 + K = 0

or K = –8

∴ Required equation of the line is 2x + 3y = 8.

10.7 STRAIGHT LINE PERPENDICULAR TO THE GIVEN LINE

Ax + By + C = 0

Let A1

x + B1y + C

1 = 0

be any line perpendicular to the given line ... (i)

Ax + By + C = 0

Condition for perpendicularity of lines (i) and (ii) is ... (ii)

AA1 + BB

1 = 0

⇒A

B

B

AK1 1

1=– = (say)

MATHEMATICS

Notes


368

Straight lines

⇒ A1 = BK

1 and B

1 = –AK

1

With these values of A1 and B

1, (i) gives

Bx – Ay +C

K1

1

0= = 0

or Bx – Ay + K = 0 where KC

K= 1

1... (iii)

Hence, the line (iii) is perpendicular to the given line (ii)

We observe that in order to get a line perpendicular to the given line we have to follow thefollowing procedure :

(i) Interchange the coefficients of x and y

(ii) Change the sign of one of them.

(iii) Change the Constant term to a new constant K (say), and evaluate it from givencondition.

Example 10.34 Find the equation of the line which passes through the point (1, 2) and is

perpendicular to the line 2x + 3y + 6 = 0.

Solution : Following the procedure given above, we get the equation of line perpendicular tothe given equation as 3x – 2y + K = 0 ...(i)

(i) passes through the point (1, 2), hence

3 × 1 – 2 × 2 + K = 0 or K = 1

∴ Required equation of the straight line is 3x – 2y + 1 = 0.

Example 10.35 Find the equation of the line which passes through the point

(x2, y

2) and is perpendicular to the straight line y y

1 = 2a (x + x

1).

Solution : The given straight line is yy1 – 2ax – 2ax

1 = 0 ...(i)

Any straight line perpendicular to (i) is 2ay + xy1 + C = 0

This passes through the point (x2, y

2)

∴ 2ay2 + x

2y

1 + C = 0

⇒ C = – 2ay2 – x

2y

1

∴ Required equation of the spraight line is

2a (y – y2) + y

1 (x – x

2) = 0

MATHEMATICS 369

Notes


Straight lines


1. Find the equation of the straight line which passes through the point (0, –2) and is parallelto the straight line 3x + y = 2.

2. Find the equation of the straight line which passes through the point (–1, 0) and is parallelto the straight line y = 2x + 3.

3. Find the equation of the straight line which passes through the point (0, –3) and isperpendicular to the straight line x + y + 1 = 0.

4. Find the equation of the line which passes through the point (0, 0) and is perpendicular tothe straight line x + y = 3.

5. Find the equation of the straight line which passes through the point (2, –3) and isperpendicular to the given straight line 2a (x + 2) + 3y = 0.

6. Find the equation of the line which has x - intercept –8 and is perpendicular to the line3x + 4y – 17 = 0.

7. Find the equation of the line whose y-intercept is 2 and is parallel to the line2x – 3y + 7 = 0.

8. Prove that the equation of a straight line passing throngh (a cos3 θ, a sin3 θ) andperpendicular to the sine x sec θ + y cosec θ = a is x cos θ – y sin θ = a cos 2θ.

10.8 APPLICATION OF COORDINATE GEOMETRY

Let us take some examples to show how coordinate geometry can be gainfully used to provesome geometrical results.

Example 10.36 Prove that the diagonals of a rectangle are equal.

Solution : Let us take origin as one vertex of the rectangle without any loss of generality andtake one side along x–axis of length a and another side of length b along y–axis, as shown inFig. 10.12. Then the coordinates of the points A, B and C are (a, 0), (a, b) and (0, b) respectively

Length of diagonal AC = (a – 0) + (0 – b) = a + b2 2 2 2

Length of diagonal OB = (a – 0)2 + (0 – b)2 = a2 + b2

∴ AC = OB

i.e., the diagonals of the rectangle OA BC are equal. Thus, in general, we can say that thelength of the diagonals of a rectangle are equal.

MATHEMATICS

Notes


370

Straight lines

Fig. 10.12

O

CB( ,b)0(a,b)

(a, )0

( )0,0 AX

Y

Example 10.37 Prove that angle in a semi circle is a right angle.

Solution : Let O (0,0) be the centre of the circle and (– a,0) and (a,0) be the end- points ofone of its diameter. Let the circle intersects the y-axis at (0, a) [see Fig. 10.13]

∴ Slope of BQ =a – 0

0 – a = 1 = m1

Slope of AQ =0 – a

a – 0 = –1 = m2

As m1

m2 = – 1 ⇒ AQ is perpendicular to BQ

∴ BQ A is a right angle.

Again, let P (a cos θ, a sin θ) be a general point on the circle.

∴ Slope of PB = 1

sin sin

cos 1 cos

am

a

= =+ + (say)

Fig. 10.13

O

Q

B

( ,a)0

A(a, )0( )0,0

A

B(a, )0

Y

P( cos , sin )a aθ θ

MATHEMATICS 371

Notes


Straight lines

Slope of QA = 2

0 sin sin

cos 1 cos

am

a a

− = =− − (say)

and m1. m

2 =

sin

cos– sin

– cos

1 1+

FHGIKJFHG

IKJ

2 2

2 2

sin sin1

1 cos 1 cos

= − = − = −− −

∴ P A is perpendicular to PB.

∠BPA is a right angle.

Example 10.38 Medians drawn from two vertices to equal sides of an isosceles triangle are

equal.

Solution : Let OAB be an equilateral triangle with vertices (0,0), (a,0) and ,2

aa

Let C and D be the mid-points of AB and OB respectively.

Fig. 10.14

C

O

D

A(a, )0(0,0)x

y ,2

aB a

∴ C has coordinates3

4 2

a a,FHGIKJ and

D has coordinatesa a

4 2,FHGIKJ

∴ Length of OC =3

4 2

9

16 4 413

22

2 2a a a a aFHGIKJ + = + =( )

and Length of AD = (a – a

4) + (0 – a

2) =

a

4132 2

∴ Length of OC = length of AD

Hence the proof.

MATHEMATICS

Notes


372

Straight lines

Example 10.39 The line segment joining the mid-points of two sides of a triangle is parallel

to the third side and is half of it.

Solution : Let the coordinates of the vertices be O (0, 0), A (a, 0) and B (b,.c). Let C and Dbe the mid-points of OB and AB respectively

∴ The coordinates of C areb c

2 2,FHGIKJ

The coordinates of D area b c+FHGIKJ2 2

,

∴ Slope of CD =

c c

a b b2 2

2 2

–

–+ = 0

Slope of OA, i.e, x – axis = 0

∴ CD is parallel to OA

Length of CD =2 2

2 2 2 2 2

a b b c c a+ − + − =

Length of OA = a

∴ CD =1

2 OA

which proves the above result.

Example 10.40 If the diagonals of a quadrilateral are perpendicular and bisect each other,

then the quadrilateral is a rhombus.

Solution : Let the diagonals be represented along OXand OY as shown in Fig. 10.16 and suppose that thelengths of their diagonals be 2a and 2b respectively.

OA = OC = a and OD = OB = b

AD = 2 2 2 2OA OD a b+ = +

AB = OB OA a b2 2 2 2+ = +

Fig. 10.15

C

O

D

x

y

B(b,c)

( ,0)A a

(0,0)

Fig. 10.16

OC

B

X

Y

D

(0, )b−

( ),0a

( )0,b

( ),0A a−

MATHEMATICS 373

Notes


Straight lines

Similarly BC = a b2 2+ = CD

As AB = BC = CD = DA

∴ A BCD is a rhumbus.


Using coordinate geometry, prove the following geometrical results:

1. If two medians of a triangle are equal, then the triangle is isosceles.

2. The diogonals of a square are equal and perpendicular to each other.

3. If the diagonals of a parallelogram are equal, then the paralle logram is a rectangle.

4. If D is the mid-point of base BC of ∆ ABC, then AB2 + AC2 = 2 (AD2 + BD2)

5. In a triangle, if a line is drawn parallel to one side, it divides the other two sidesproportionally.

6. If two sides of a triangle are unequal, the greater side has a greater angle opposite to it.

7. The sum of any two sides of a triangle is greater than the third side.

8. In a triangle, the medians pass throngh the same point.

9. The diagonal of a paralleogram divides it into two triangles of equal area.

10. In a parallelogram, the pairs of opposite sides are of equal length.

LET US SUM UP

• The equation of a line parallel to y-axis is x = a and parallel to x-axis is y = b.

• The equation of the line which cuts off intercept c on y-axis and having slope m isy = mx + c

• The equation of the line passing through A(x1, y

1) and having the slope m is

y – y1 = m (x – x

1)

• The equation of the line passing through two points A(x, y1) and B(x

2, y

2) is

y – y1 =

y y

x x2 1

2 1

–– (x – x

1)

• The equation of the line which cuts off interceptsa andb onx−axis andy– axis respectively

isx

a

y

b+ = 1

• The equation of the line in normal or perpendicular form isx cos + y sin = p

MATHEMATICS

Notes


374

Straight lines

where p is the length of perpendicular from the origin to the line and is the angle whichthis perpendicular makes with the positive direction of thex-axis.

• The equation of a line in the parametric form is 1 1

cos sin

x x y yr

− −= =

where r is the distance of any point P (x, y) on the line from a given point A(x1, y

1) on the

given line and is the angle which the line makes with the positive direction of thex-axis.

• The general equation of first degree inx and y always represents a straight line providedA and B are not both zero simultaneously.

• From general equation Ax + By + C = 0 we can evaluate the following :

(i) Slope of the line = – A

B(ii) x-intercept = – C

A (iii) y-intercept = – C

B

(iv) Length of perpendicular from the origin to the line =| |

( )

C

A B2 2+

(v) Inclination of the perpendicular from the origin is given by

cos =A

A B( )2 2+ ; sin =B

(A + B )2 2

where the upper sign is taken for C > 0 and the lower sign for C < 0; but if C = 0 theneither only the upper sign or only the lower sign are taken.

• If be the angle between two lines with slopes m1m

2 then

1 2

1 2

tan1

m m

m m −=

+

(i) Lines are parallel if m1 = m

2. (ii) Lines are perpendicular if m

1m

2 = –1.

(B) Angle between two lines of general form is:

tan θ =A B A B

A A B B2 1 1 2

1 2 1 2

–+

(i) Lines are parallel if A1B

2 – A

2B

1 = 0 or

A B

A B1 1

2 2 = 0

(ii) Lines are perpendicular if A1A

2 + B

1B

2 = 0.

• Distance of a given point (x1,y

1) from a given lineAx + By + C =0 is d =

Ax By C

A B1 1

2 2

+ ++( )

• Equation of a line parallel to the line Ax + By + C = 0 is Ax + By + k = 0

• Equation of a line perpendicular to the line Ax + By + C = 0 is Bx – Ay + k = 0

MATHEMATICS 375

Notes


Straight lines

SUPPORTIVE WEB SITES

http://www.wikipedia.org

http://mathworld.wolfram.com

1. Find the equation of the straight line whosey-intercept is –3 and which is:

(a) parallel to the line joining the points (–2,3) and (4, –5).

(b) perpendicular to the line joining the points (0, –5) and (–1, 3).

2. Find the equation of the line passing through the point (4, –5) and

(a) parallel to the line joining the points (3, 7) and (–2, 4).

(b) perpendicular to the line joining the points (–1, 2) and (4, 6).

3. Show that the points (a, 0), (0, b) and (3a, – 2b) are collinear. Also find the equation ofthe line containing them.

4. A (1, 4), B (2, –3) and C (–1, –2) are the vertices of triangle ABC. Find

(a) the equation of the median through A.

(b) the equation of the altitude througle A.

(c) the right bisector of the side BC.

5. A straight line is drawn through point A (2, 1) making an angle of6

with the positive

direction of x-axis. Find the equation of the line.

6. A straight line passes through the point (2, 3) and is parallel to the line 2x + 3y + 7 = 0.Find its equation.

7. Find the equation of the line havinga and b as x-intercept and y-intercepts respectively.

8. Find the angle between the lines y = (2 – 3 ) x + 5 and y = (2 + 3 ) x – d.

9. Find the angle between the lines 2x + 3y = 4 and 3x – 2y = 7

10. Find the length of the per pendicular drawn from the point (3, 4) on the straight line12 (x + 6) = 5 (y –2).

11. Find the length of the perpendicula from (0, 1) on 3x + 4 y + 5 = 0.

12. Find the distance between the lines

2x + 3y = 4 and 4x + 6y = 20

TERMINAL EXERCISE

http://www.wikipedia.org

http://mathworld.wolfram.com

MATHEMATICS

Notes


376

Straight lines

13. Find the length of the perpendicular drawn from the point (–3, – 4) on the line4x – 3y = 7.

14. Show that the product of the perpendiculars drawn from the points on the straight line

x

a

y

bcos + sin θ = 1 is b2.

15. Prove that the equation of the straight line which passes through the point (a cos3 θ, bsin3 θ) and is perpendicular to x secθ + y cosec θ = a is x cosθ - y cosec θ = a cos 2θ

16. Prove the following geometrical results :

(a) The altitudes of a triangle are concurrent.

(b) The medians of an equaliateral triangle are equal.

(c) The area of an equalateral triangle of side a is3

4

2a

(d) If G is the centroid of ∆ ABC, and O is any point, then OA2 + OB2 + OC2 =GA2 + GB2 + CG2 + 3OG2.

(e) The perpendicular drawn from the centre of a circle to a chord bisects the chord.

MATHEMATICS 377

Notes


Straight lines

ANSWERS


1. Straight line 2. (a) 1 (b) –1 (c) 1

3. 7 4. 20 5. (a) y = –4 (b) x = –3

6. x = 5 7. y + 7 = 0


1. (a) y = 2x –2 (b) Slope =–4

,3

y – intercept = 2

2. 3y = –3x–1 3. Slope =1

2, y-intercept = –2

4. 3x + 7y = 7 5. y = x + 1; x + y – 3 = 0 6. 3x –2y = 0

7. (a) x + y = –1 (b) Equation of the diagonal AC = 2x – y – 4 = 0

Equation of the diagonal BD = 2x – 11y + 66 = 0

8. x – 2 = 0, x – 3y + 6 = 9 and 5x – 3y – 2 = 0

9. 2x + 3y = 6 10. 3x + y = 6 11. 3x + 4y = 1 12. x + y = 2 2

14.

x y− = −21

2

11

2 and the co-ordinates of the point are

1 12 ,1

2 2

+ +


1. A and B are not both simltaneously zero 2. y x= –2 –5

3

5

3. (a) x –intercept =–c

m; y-intercept = c (b) x-intercept =

–83

; y-intercept =8

3

(c) x-intercept = –4; y-intercept = 6

4. 2 13 units 5. 1sec cos

x x

p p ec + =

MATHEMATICS

Notes


378

Straight lines

6. (a)–35

4

5x + y – 2 = 0 (b)

–35

4

5x + y = 0

7. The first line is nearer from the origin.


1. (a) m1 =

– A

B(b) Am

1 = B

2. θ = tan− FHGIKJ

1 7

6

3. (a) (i)A

B

A

B1

1

2

2

= (ii) A1 A

2 + B

1 B

2 = 0

(b) Parallel (c) Perpendicular.


1. d =16

132. 0 2 2, ( )

b

aa a b± +F

HGIKJ 3. 0

32

3,FHGIKJ

4.14

585.

3

44 5 0( ),±FHGIKJ


1. 3x + y + 2 = 0 2. y = 2x + 2 3. x – y = 3 4. y = x

5. 3x – 2ay = 6 (a – 1) 6. 4x – 3y + 32 = 0 7. 2x – 3y + 6 = 0

TERMINAL EXERCISE

1. (a) 4x + 3y + 9 = 0 (b) x – 8y – 24 = 0

2. (a) 3x – 5y – 37 = 0 (b) 5x – 8y – 60 = 0

4. (a) 13x – y – 9 = 0 (b) 3x – y + 1 = 0

(c) 3x – y – 4 = 0

5. x – 3y = 2 – 3 6. 2x + 3y +13 = 0

7. bx + ay = ab 8.2

9.

2

10.

98

13

11.9

512.

6

1313.

7

5

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